Problem: Complete the equation. $\dfrac{3}{4}+ \dfrac{3}{4}+ \dfrac{3}{4} = 9 \times$
Explanation: Let's figure out what $\dfrac{3}{4} + \dfrac{3}{4} + \dfrac{3}{4}$ equals. $\dfrac{0}{4}$ $\dfrac{3}{4}$ $\dfrac{6}{4}$ $\dfrac{9}{4}$ $\llap{{+}}\!\frac{3}{4}$ $\llap{{+}}\!\frac{3}{4}$ $\llap{{+}}\!\frac{3}{4}$ $\dfrac{3}{4} + \dfrac{3}{4} + \dfrac{3}{4}= \dfrac{9}{4}$ ${\text{What number}}$ can we add $9$ times to make $\dfrac94$ ? $\dfrac{0}{4}$ $\dfrac{1}{4}$ $\dfrac{2}{4}$ $\dfrac{3}{4}$ $\dfrac{4}{4}$ $\dfrac{5}{4}$ $\dfrac{6}{4}$ $\dfrac{7}{4}$ $\dfrac{8}{4}$ $\dfrac{9}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $=\overbrace{{\dfrac1{4}} +{\dfrac1{4}} +{\dfrac1{4}} + {\dfrac1{4}} +{\dfrac1{4}} +{\dfrac1{4}} + {\dfrac1{4}} + {\dfrac1{4}} + {\dfrac1{4}}}^{{9}\text{ fourths}} $ $=\dfrac{{9}\times{1}}{{4}}$ $\dfrac34 + \dfrac34 + \dfrac34 = 9 \times {\dfrac14}$